Question

यदि
$x + \frac{1}{x} = 7 \: then \: {x}^{3} + \frac{1}{ {x}^{3} }$
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Answer 1

322

Step-by-step explanation:

we know that:

(a + b)3 = a3 + b3 +3ab(a+b)

$(x + \frac{1}{x})^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) \\ (x + \frac{1}{x})^{3} = {x}^{3} + \frac{1}{x^3} + 3(x + \frac{1}{x})...(1)\\put \\ x + \frac{1}{x} = 7 in equation..(1)\\ {7}^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 7 \\ 343 = {x}^{3} + \frac{1}{ {x}^{3} } + 21 \\ {x}^{3} + \frac{1}{ {x}^{3} } = 343 - 21 \\ {x}^{3} + \frac{1}{ {x}^{3} } = 322$

I hope this help you

Answer 2

Answer:

$\large\boxed{\sf{322}}$

Step-by-step explanation:

Given that,

$x + \dfrac{1}{x} = 7 \: \: \: \: \: \: ............(1)$

Squaring both the sides, we get,

$= > {(x + \dfrac{1}{x} )}^{2} = {7}^{2} \\ \\ = > {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 49 \\ \\ = > {x}^{2} + \dfrac{1}{ {x}^{2} } - 1 = 46 \: \: \: \: \: .......(2)$

Now, to find the value of,

${x}^{3} + \dfrac{1}{ {x}^{3} }$

We know that,

${x}^{3} + \dfrac{1}{ {x}^{3} } = (x + \dfrac{1}{x} )( {x}^{2} - 1 + \dfrac{1}{ {x}^{2} } )$

Now, substitute the values from (1) and (2),

Therefore we will get,

$= > {x}^{3} + \dfrac{1}{ {x}^{3} } = 7 \times 46 \\ \\ = > {x}^{3} + \dfrac{1}{ {x}^{3} } = 322$

Hence, the required Answer is 322.